Download Exam 1 Sol

APPM 1350
Exam 1
1. Consider the function h(x) = √
Fall 2015
4x
− 25
x2
(a) (4 points) Give the domain of this function in interval notation.
(b) (6 points) Use the appropriate limits to identify any vertical asymptotes. If none exist, write “None”
and explain why.
(c) (6 points) Use the appropriate limits to identify any horizontal asymptotes. If none exist, write
“None” and explain why.
(d) (4 points) Is the function’s symmetry even, odd, or neither?
Solution:
(a) The expression under the radical needs to be positive: x2 − 25 > 0 ⇒ x < −5 or x > 5. The
domain is (−∞, −5) ∪ (5, ∞) .
(b) There are possible vertical asymptotes where the denominator equals 0 at x = −5, 5.
√
4x
lim √
= ∞ since 4x → 20 and x2 − 25 → 0 with positive values.
+
2
x→5
x − 25
√
4x
lim √
= −∞ since 4x → −20 and x2 − 25 → 0 with positive values.
x→−5−
x2 − 25
There are vertical asymptotes at x = 5 and x = −5 .
Grading Comments: Observe that the limit of h(x) is undefined as x → 5− and x → −5+ .)
√
√
(c) Note that x2 = x for x ≥ 0 and x2 = −x for x < 0.
4
4x
4x
4Z
x
= lim √ p
= lim p
=√
=4
2
2
2
x→∞
x→∞
1−0
− 25
x 1 − 25/x
x 1 − 25/x
Z
4x
4x
4Z
x
−4
p
= −4
lim √
= lim √ p
= lim
=√
2
2
2
2
x→−∞
x→−∞
x→−∞
1−0
x − 25
−Z
x 1 − 25/x
x 1 − 25/x
lim √
x→∞
x2
There are horizontal asymptotes at y = 4 and y = −4 .
(d) Since h(−x) = √
−4x
= −h(x), then h is odd .
x2 − 25
2. (21 points, 7 points each) Evaluate the following limits
3
3
(a) lim
− 2
t→0 t
t +t
| sin x|
(b) lim
x→0 sin x
sin θ
(c) lim
θ→0 θ + tan θ
Solution:
3(t + 1) − 3
3
3tC
3
= lim
(a) (HW 1.4.24) lim
− 2
= lim
= 3
t→0
t→0 t
t→0 tC (t + 1)
t +t
t(t + 1)
(
sin x,
0≤x≤π
(b) Note that |sin x| =
− sin x, −π ≤ x ≤ 0.
| sin x|
sin x
lim
= lim
=1
x→0+ sin x
x→0+ sin x
| sin x|
− sin x
lim
= lim
= −1
−
−
sin x
sin x
x→0
x→0
| sin x|
is undefined since the left-hand limit and right-hand limit are not equal.
lim
x→0 sin x
sin θ
1
sin θ
sin θ
1
1
θ
θ
(c) (HW 1.4.55) lim
=
lim
= lim
·
=
=
1
sin
θ
sin
θ
1
θ→0 θ +
θ→0 θ + tan θ
θ→0 1 +
1+1·1
2
θ
cos θ
θ · cos θ
sin θ
sin θ
sin θ
does not equal
+
. It is true that
θ + tan θ
θ
tan θ
a
a+b
a b
a a
= + . It is not true that
equals + .
c
c c
b+c
b
c
√
√
x− 5
3. Consider the function g(x) = 2
.
x − 6x + 5
Grading Comments: The fraction
(a) (6 points) Give the domain of this function in interval notation.
(b) (6 points) Evaluate lim g(x).
x→5
(c) (8 points) The function g has a removable discontinuity at x = a. The discontinuity can be removed
by creating a new function h(x).
(
g(x) x 6= a
h(x) =
b
x=a
Use the definition of continuity of a function to find the values of the constants a and b.
Solution:
√
(a) The domain of x is [0, ∞). The function g is undefined when the denominator x2 − 6x + 5 =
(x − 1)(x − 5) equals zero at x = 1, 5. The domain is therefore [0, 1) ∪ (1, 5) ∪ (5, ∞) .
Grading Comments: When finding the domain of the function, the square root expression must be
considered. For this problem that means x cannot be negative.
√
√
√
√
X
XX
x− 5
x+ 5
x−
5
1
1
√ = lim
√ =
√ = √
(b) lim 2
·√
√
X
x→5 x − 6x + 5
x + 5 x→5 (x − 1)(xX
−X5)
4(2 5)
8 5
X x+ 5
(c) Since g(5) is undefined and lim g(x) exists, the function g has a removable discontinuity at x = 5.
x→5
1
The function h is continuous at a if h(a) = lim h(x). Let a = 5, b = √ . Then
x→a
8 5
1
h(5) = lim h(x) = lim g(x) = √ .
x→5
x→5
8 5
Grading Comments: Note that the problem asked for the definition of continuity. A solution that
lacked the definition of continuity was considered incomplete.
4. (15 points) Let g(x) =
x−1
.
x+1
(a) Use the definition of derivative to find the slope of the tangent line to y = g(x) at x = 0.
(b) Find the equation of the tangent line to y = g(x) at x = 0.
(c) Find the equation of the normal line to y = g(x) at x = 0.
Solution:
(a)
g(a + h) − g(a)
h
g(h) − g(0)
g 0 (0) = lim
= lim
h→0
h→0
h
2S
h
= lim
= 2
h→0 S
h(h + 1)
g 0 (a) = lim
h→0
h−1
h+1
− (−1)
1 h − 1A + h + 1A
= lim ·
h→0 h
h
h+1
Alternate Solution.
g(a + h) − g(a)
h→0
h
a+h−1
− a−1
= lim a+h+1 a+1
h→0
h
1 (a + h − 1)(a + 1) − (a − 1)(a + h + 1)
= lim ·
h→0 h
(a + h + 1)(a + 1)
g 0 (a) = lim
2 +Z
2 − a +Z
1 (@
a@
ah
a@
A) − (@
A)
Z− a
Z− h + a
A+a
A+h−1
A ah
A−1
·
h→0 h
(a + h + 1)(a + 1)
2S
h
1
= lim ·
h→0 S
h (a + h + 1)(a + 1)
2
=
(a + 1)2
= lim
g 0 (0) = 2
(b) At x = 0, y(0) = −1 and y 0 (0) = 2. Using point-slope form, the tangent line is
y − y1 = m(x − x1 ) ⇒ y + 1 = 2x or y = 2x − 1.
(c) The normal line passes through the same point and has slope −1/2. The equation of the normal
line is
1
1
y + 1 = − x or y = − x − 1.
2
2
5. (24 points, 6 points each) Some unrelated, short answer questions.
(1 + h)10 − 1
represents the derivative of some function f at some
h→0
h
(a) (HW 2.1.31) The limit lim
number a.
(i) Find a function f and a value for a. (ii) What is the value of the limit?
(b) Does x + tan x = 1 have a solution? Justify your answer.
(c) Sometimes a function f is not continuous on its domain but |f | is continuous, on the same domain.
Find an example of such a function f (i.e. f is not continuous at a point in its domain but |f | is).
Either sketch the graph of both |f | and f or find a formula that illustrates this.
(d) A factory manufactures metal cubes of volume V = 8000 cm3 . An error tolerance of ±5 cm3 is
allowed, which corresponds to a side length s between 19.996 and 20.004 cm. In terms of the formal
definition of lim f (x) = L, identify x, a, f (x), L, δ, and . No further explanation is necessary for
x→a
this problem.
Solution:
f (a + h) − f (a)
(1 + h)10 − 1
= lim
. One possible solution is
h→0
h→0
h
h
(a) (HW 2.1.31) (i) f 0 (a) = lim
f (x) = x10 and a = 1 .
(ii) Since f 0 (x) = 10x9 , the limit value equals f 0 (1) = 10 .
(b) Let f (x) = x + tan x − 1. Then f (0) = 0 + 0 − 1 = −1 < 0 and f ( π4 ) = π4 + 1 − 1 = π4 > 0.
By the Intermediate Value
Theorem, since f is continuous in the interval [0, π4 ], there is a value of
π
c in the interval 0, 4 such that f (c) = 0.
Grading comments: The domain of tan(x) is not R. Indeed, tan x is not defined for π/2 + nπ,
with n an integer. Note that the IVT can be applied only on a closed interval included in the domain
of the function. The function must be continuous over the whole interval: hence it is not correct to
use the IVT on [0, π] as this interval contains π2 , where tan x is discontinuous.
(c) Below are two possible solutions.
�  � (�)
�   � (�)
�
�
�
�
�
�
-�
-�
�  � (�)
�   � (�)
�
�
�
�
�
-�
-�
�
-�
-�
Grading comments: Note that the graph of a piecewise function must clearly indicate what the
value of the function is at any jump discontinuity by using clearly distinguishable open and closed
circles. Two closed circles corresponding to one value of x does not result in a function.
(d) The formal definition of limit states that lim f (x) = L if for every > 0 there is a δ > 0 such
x→a
that if 0 < |x − a| < δ then |f (x) − L| < . For this problem, the function is V (s) = s3 . Since
V (20) = 8000, the corresponding limit is lim V (s) = 8000 so x = s, a = 20, f (x) = V (s) , and
s→20
L = 8000 . For = 5 , the value of δ is |20.004 − 20| = |20 − 19.996| or δ = 0.004 .
Grading comments: The values for ε and δ are positive quantities, by definition. Do not include
any ± symbol.